Question: $g(x) = -6x^{2}+5(h(x))$ $h(n) = -4n^{2}$ $f(t) = -t^{3}+t^{2}+6t-3(h(t))$ $ g(f(0)) = {?} $
Explanation: First, let's solve for the value of the inner function, $f(0)$ . Then we'll know what to plug into the outer function. $f(0) = -0^{3}+0^{2}+(6)(0)-3(h(0))$ To solve for the value of $f$ , we need to solve for the value of $h(0)$ $h(0) = -4(0^{2})$ $h(0) = 0$ That means $f(0) = -0^{3}+0^{2}+(6)(0)+(-3)(0)$ $f(0) = 0$ Now we know that $f(0) = 0$ . Let's solve for $g(f(0))$ , which is $g(0)$ $g(0) = -6(0^{2})+5(h(0))$ To solve for the value of $g$ , we need to solve for the value of $h(0)$ $h(0) = -4(0^{2})$ $h(0) = 0$ That means $g(0) = -6(0^{2})+(5)(0)$ $g(0) = 0$